Question

Verify that the point P(a cos θ, b sin θ) lies on the ellipse
x
2
a
2
+
y
2
b
2
= 1,
where a and b are the semi-major and semi-minor axes respectively of the ellipse . Find the
gradient of the tangent to the curve at P and show that the equation of the normal at P is
ax sin θ − by cos θ = (a
2 − b
2
) sin θ cos θ.
If P is not on the axes and if the normal at P passes through the point B(0, b), Show that
a
2 > 2b
2
. If further, the tangent at P meets the y-axis at Q, show that
|BQ| =
a
2
b
2
.

Answer 1

If the point $P(acos\theta,bsin\theta)$ satisfies the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , then it lies on the given ellipse .

Now $\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{(acos\theta)^2}{a^2}+\frac{(bsin\theta)^2}{b^2}$

$= \frac{a^2cos^2\theta}{a^2}+\frac{b^2sin^2\theta}{b^2}$

$= cos^2\theta+sin^2\theta$

$=1.$

$\therefore$ point $P(acos\theta,bsin\theta)$ lies on the given ellipse. (PROVED).

Now the equation of the tangent at the point $P(acos\theta,bsin\theta)$ is given by

$\frac{x*acos\theta}{a^2}+\frac{y*bsin\theta}{b^2}=1\\or, \space \frac{xcos\theta}{a}+\frac{ysin\theta}{b}=1\\or, \space \frac{bxcos\theta+aysin\theta}{ab}=1\\or, \space bxcos\theta+aysin\theta=ab\\or, \space aysin\theta=-bxcos\theta+ab\\or, \space y=-\frac{bxcos\theta}{asin\theta}+\frac{ab}{asin\theta}\\or, \space y=-(\frac{b}{a}tan\theta)x +bcosec\theta.$

$\therefore$ Gradient of the tangent at the point $P(acos\theta,bsin\theta)$ is $-(\frac{b}{a}tan\theta)$ . Answer.

Equation of Normal: We know that normal is perpendicular to the tangent..

so if equation of tangent is $bxcos\theta+aysin\theta=ab$ then equation of normal is of the form

$axsin\theta-bycos\theta=k, \space where \space k \space is \space constant.$

Now as this normal is passing through $P(acos\theta,bsin\theta)$ ...by putting this coordinate into the equation of normal we get the value of $k$ .

$\therefore a*acos\theta*sin\theta-b*bsin\theta*cos\theta=k\\or, \space k=a^2sin\theta cos\theta-b^2sin\theta cos\theta\\or, \space k= (a^2-b^2)sin\theta cos\theta$

$\therefore$ The equation of normal at the point $P(acos\theta,bsin\theta)$ is given by

$axsin\theta-bycos\theta= (a^2-b^2)sin\theta cos\theta.$ PROVED.

Now if the equation of normal is passing through B(0,b) then we have

$a*0*sin\theta-b*b*cos\theta=(a^2-b^2)sin\theta cos\theta\\or, \space -b^2cos\theta=(a^2-b^2)sin\theta cos\theta.\\ or, \space -b^2=(a^2-b^2)sin\theta \\or, \space sin\theta = \frac{-b^2}{a^2-b^2}$

Now as we know $-1\leq sin\theta,\\$

$\therefore -1\leq \frac{-b^2}{a^2-b^2}\\or, 1\geq \frac{b^2}{a^2-b^2}\\ or, a^2-b^2 \geq b^2\\or, a^2 \geq2b^2$ PROVED.

Now tangent at $P(acos\theta,bsin\theta)$ which is $bxcos\theta+aysin\theta=ab$ , meets the Y axis at Q.

putting x=0 ,we get

$0+aysin\theta=ab\\or, \space ysin\theta=b\\or, \space y=\frac{ b}{sin\theta}.\\$

$\therefore$ The coordinate of Q is $(0,\frac{b}{sin\theta})$ .

$BQ= \bigg(\frac{b}{sin\theta}-b\bigg)\\BQ=\bigg(\frac{b}{\frac{-b^2}{a^2-b^2}}-b\bigg)$

$BQ= b(\frac{b^2-a^2}{b^2}-1)\\BQ= b*(\frac{b^2-a^2-b^2}{b^2})\\BQ=\frac{-a^2}{b}$

$\therefore |BQ|= \frac{a^2}{b}$ PROVED.

After solving question i found that BQ is $\frac{a^2}{b}$ but in a given question it was given as $\frac{a^2}{b^2}$ .. This is a mistake.